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I was looking at the question Is $7$ the only prime followed by a cube? and I was running through some of the low powers of $n^m-1$

I tried $m=2$ and got that $n^2-1=(n+1)(n-1)$ I reasoned that $n+1>1$ because otherwise $n-1<0$ and cannot be prime. So setting $n-1=1$ we get $n=2$ and that gives us our prime of $2+1=3$ which should be the largest prime followed by a square.

Next I tried $m=4$ and got that $n^4-1=(n^2+1)(n^2-1)$ with $n^2+1>1$. But now $n^2-1=1$ implying that $n = \pm \sqrt 2$ which is not an integer. I took this as a contradiction and reason that there are no primes following powers of 4.

Finally, I looked at $m=2k$ and got that $n^{2k}-1=(n^k+1)(n^k-1)$ with $n^k+1>1$. I get the same thing $n^k-1=1$ implying that $n = \pm \sqrt[k] 2$ which is not an integer for $k>1$ (which gives a reason that $m=2$ works).

Is my reasoning correct? Are there really no primes following even powers greater than 2 of n?

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You should probably say even powers following primes: The primes come before the even powers in this case. But yes, your reasoning is correct. Note that $n^k - 1$ for $k > 1$ can always be decomposed into a further factorization, at the very least being $(n-1)(n^{k-1} + n^{k-2} + n^{k-3} + \dots + 1)$ when $k$ is prime –  Soke Feb 17 at 16:14
    
Edited to correct, thanks for pointing that out. –  kleineg Feb 17 at 16:17
    
Any even power of $n$ is also a square... –  TonyK Feb 17 at 16:18
    
Yes $n^{2k}$ is a square, but $n^{2k}-1$ is not a square. –  kleineg Feb 17 at 16:21
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Maybe, you have got a very good observation, and it may help me with my research. –  Hawk Feb 17 at 16:25

1 Answer 1

up vote 2 down vote accepted

Yes, your reasoning is correct. If $n^{2k}-1$ is prime then $n^k-1=1$ and hence $n^k=2^1.$

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